An a.c. circuit contains resistance of $12 \Omega$ and inductor of inductive reactance $5 \Omega$. The phase angle between current and potential difference will be

The phase angle $\phi$ between current and potential difference in an AC circuit with resistance $R = 12\,\Omega$ and inductive reactance $X_L = 5\,\Omega$ satisfies $\tan \phi = \frac{X_L}{R} = \frac{5}{12}$.

The impedance is $Z = \sqrt{R^2 + X_L^2} = \sqrt{144 + 25} = 13\,\Omega$.

From the impedance triangle, $\cos \phi = \frac{R}{Z} = \frac{12}{13}$ and $\sin \phi = \frac{X_L}{Z} = \frac{5}{13}$.

Therefore, $\phi = \cos^{-1}\left(\frac{12}{13}\right)$ or equivalently $\phi = \sin^{-1}\left(\frac{5}{13}\right)$.

Comparing with the given options, the correct expression is $\cos^{-1}\left(\frac{12}{13}\right)$ which corresponds to option D.